Solve the Equation for Solutions in the Interval [0, 2 $$\sin ^ { 2 } 2 x = 1$$

Solve the equation for solutions in the interval [0, 2 π) .
- $$\sin ^ { 2 } 2 x = 1$$

A) $\left\{ 0 , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 } \right\}$
B) $\varnothing$
C) $\left\{ \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 } \right\}$
D) $\left\{ \frac { \pi } { 8 } , \frac { 9 \pi } { 8 } \right\}$

Correct Answer:

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